Question

The vertices of a quadrilateral ABCD are A(6, – 2), B(9, 2), C(5, – 1) and D(2, – 5). Prove that ABCD is a rhombus, and not a square.

Answer 1

Step 1: Find the lengths of the sides using the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \] Side AB: \[ AB = \sqrt{(9 - 6)^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5. \] Side BC: \[ BC = \sqrt{(5 - 9)^2 + (-1 - 2)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = 5. \] Side CD: \[ CD = \sqrt{(2 - 5)^2 + (-5 - (-1))^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = 5. \] Side DA: \[ DA = \sqrt{(6 - 2)^2 + (-2 - (-5))^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5. \]  All sides are equal, so $ABCD$ is a rhombus. Step 2: Check diagonals to confirm it is not a square. Diagonal AC: \[ AC = \sqrt{(5 - 6)^2 + (-1 - (-2))^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}. \] Diagonal BD: \[ BD = \sqrt{(2 - 9)^2 + (-5 - 2)^2} = \sqrt{(-7)^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2}. \]  Since the diagonals are not equal ($AC \neq BD$), $ABCD$ is not a square. Correct Answer: $ABCD$ is a rhombus, but not a square.